Question: In a right prism with triangular bases, given the sum of the areas of three mutually adjacent faces (that is, of two lateral faces and one base) is 24, find the maximum volume of the prism.

[asy]
unitsize(1 cm);

pair A, B, C, D, E, F;

A = (0,0);
B = (3,-1);
C = (-1,-2);
D = A + (0,-4);
E = B + (0,-4);
F = C + (0,-4);

draw(A--B--C--cycle);
draw(E--F);
draw(F--D--E,dashed);
draw(A--D,dashed);
draw(B--E);
draw(C--F);
[/asy]
Solution: Let the base triangles have sides $a$ and $b$ with included angle $\theta,$ and let the right prism have altitude $h$.

[asy]
unitsize(1 cm);

pair A, B, C, D, E, F;

A = (0,0);
B = (3,-1);
C = (-1,-2);
D = A + (0,-4);
E = B + (0,-4);
F = C + (0,-4);

draw(A--B--C--cycle);
draw(E--F);
draw(F--D--E,dashed);
draw(A--D,dashed);
draw(B--E);
draw(C--F);

label("$a$", (B + C)/2, S);
label("$b$", (A + C)/2, NW);
label("$h$", (C + F)/2, W);
label("$\theta$", C + (0.4,0.4));
[/asy]

Then the surface area constraint is

$$ah + bh + \frac12 ab \sin \theta = 24,$$and the volume is

$$V = \frac12 abh \sin \theta.$$Let $X = ah, Y = bh, Z = (ab \sin \theta) / 2$ be the areas of the three faces.  Then $X + Y + Z = 24$, and
\[XYZ = \frac{1}{2} a^2 b^2 h^2 \sin \theta = \frac{2}{\sin \theta} \left( \frac{1}{2} abh \sin \theta \right)^2 = \frac{2V^2}{\sin \theta}.\]Now the AM-GM inequality yields

$$(XYZ)^{1/3} \leq \frac{X+Y+Z}{3} = 8,$$so $XYZ \le 512$.  But
\[\frac{2V^2}{\sin \theta} = XYZ \le 512,\]so
\[V^2 \le 256 \sin \theta \le 256,\]which means $V \le 16$.

Equality occurs for $a = b = 4$, $h = 2$, and $\theta = \pi/2$, so the maximum volume of the prism is $\boxed{16}$.